Question Number 219356 by SdC355 last updated on 23/Apr/25
$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \left({z}\right)}{{z}^{\mathrm{2}} }{e}^{−{zt}} \:\mathrm{d}{z} \\ $$
Answered by breniam last updated on 23/Apr/25
$$=−\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{1}}{{z}}\right)'\left(\mathrm{sin}^{\mathrm{2}} \left({z}\right)\right)\mathrm{d}{z}=−\mathrm{0}+\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{sin}\left({z}\right)\mathrm{cos}\:\left({z}\right)}{{z}}\mathrm{d}{z}= \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{sin}\left(\mathrm{2}{z}\right)}{{z}}\mathrm{d}{z}=\left\{\overset{−} {{z}}=\mathrm{2}{z}\right\}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{sin}\left({z}\right)}{{z}}\mathrm{d}{z}=\frac{\pi}{\mathrm{2}} \\ $$